C OMPOSITIO M ATHEMATICA
V AGN L UNDSGAARD H ANSEN On the fundamental group of a mapping space. An example
Compositio Mathematica, tome 28, n
o1 (1974), p. 33-36
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33
ON THE FUNDAMENTAL GROUP OF A MAPPING SPACE.
AN EXAMPLE
Vagn Lundsgaard
HansenNoordhoff International Publishing Printed in the Netherlands
1. Introduction
The purpose of this paper is to
provide
anexample
of amapping
space between
connected, compact polyhedra having finitely generated homotopy
groups in alldimensions,
such that at least onecomponent
in the
mapping
space hasinfinitely generated
fundamental group.Throughout
K is aconnected, compact polyhedron. Mostly
K =Si,
the unit circle. For any
connected, topological
spaceX, XK
shall denote the space of continuous maps of K into Xequipped
with thecompact-
opentopology.
Inparticular XS1
denotes the freeloop
space on X. Weadopt
theterminology
that anaspherical
space is aconnected, compact
polyhedron X,
which is also anEilenberg-MacLane
spaceof type (03C0, 1),
i.e.
03C01(X) =
x and7ri(X)
= 0 for all i~
2.THEOREM: There exists an
aspherical
space X withfinitely generated fundamental
group, such thatthe free loop
spaceXS1
contains a componentwith
infinitely generated fundamental
group.A similar
phenomenon
cannothappen
inhigher
dimensions. Itis,
in
fact,
easy to prove that if X hasfinitely generated homotopy
groups indimensions ~ 2,
then all thehomotopy
groups ofXK
indimensions ~
2are also
finitely generated.
Results on the
qualitative
structure of thehomotopy
groups of amapping
space were obtained among othersby
Federer[1 J
and Thom[4].
Theexample
in the theorem above shows that Thom’s statement([4],
Theorem4)
is incorrect for the fundamental group.Finally,
I want to thank J.Eells,
A. C.Robinson,
B.Hartly
andD. B. A.
Epstein
for varioushelpful
remarksduring
thepreparation
ofthis paper.
2. Free
loop
spaces. Their fundamental groupsIn this section X is a
connected, topological
space with basepoint
*.It is well-known that the
(path-)components
in themapping
spaceXs’
can be enumerated
by xi (X, *).
The fundamental group of acomponent
inXS1 depends normally
on thecomponent
inquestion.
Todescribe how,
34
recall that the centralizer of an element g in a group G is the
subgroup
ofG defined
by Cg(G) = {g’ ~ Glg’g
=gg’l.
We have thenPROPOSITION 1:
Suppose
that1t2(X, *)
= 0 and letf : S1 ~
X be anarbitrary
basepoint preserving
maprepresenting
thehomotopy
class[f] E 7r,(X, *).
ThenPROOF: Let 03A9X denote the
ordinary loop
space onX,
i.e. the space of based maps ofSi
into Xequipped
with thecompact-open topology.
It iswell-known that the map p :
XS1 ~
X definedby
evaluation at the basepoint
ofS’,
also denoted *, is a Hurewicz fibration with fibre 03A9X. See e.g. Hu([2],
Theorem 13.1 p.83).
Consider now thefollowing part
ofthe
homotopy
sequence of thatfibration,
Since 03C01(03A9X, f) ~ 03C02(X, *)
=0, p* is a monomorphism. Hence 03C01(XS1,
f)
isisomorphic
to theimage of p*.
To determine thisimage
observe thata e xi
(X, *)
is in theimage of p*
if andonly
if there exists amap S1 x sl
~ X such that
S1 {*} ~
Xrepresents
aand {*}
xS1 ~
Xrepresents
[f].
On the other hand such a map exists if andonly
if the Whiteheadproduct
of a and[f] is
theidentity
elementof 03C01(X, *).
Since a Whiteheadproduct
of elements in a fundamental group coincides with the cor-responding
commutatorproduct,
weconclude,
that a is in theimage
ofp* if and only if 03B1[f]03B1-1[f]-1 =
1 orequivalently
that a EC[f] (03C01 (X, *)).
This proves the
proposition.
3. The
example
Let X be the
connected, compact,
2-dimensionalpolyhedron
obtainedfrom a
cylinder
over an oriented circle with basepoint by pinching
oneboundary
circle into afigure eight
and thenidentifying
theresulting
threeboundary
circlesaccording
to orientation. SeeFigure
1.Fig. 1
We shall show that this
polyhedron
satisfies therequirements
in thetheorem. This will follow
easily
fromProposition
1 andPropositions
2and 3 below.
The base
point
from the circle in thecylinder
leaves X with a basepoint.
Denote the fundamental group of X
by
03C0, and consider the elements xand y
in xgenerated by
the twoloops
indicated inFigure
1. For eachn =
0, 1, 2, ...
weput yn
=x"yx-".
Inparticular
yo = y.PROPOSITION 2: x is
generated by
the elements x and ysubject
to thesingle
relationx-1yx
=y2.
The centralizer
of
y E 03C0,Cy(03C0),
isinfinitely generated
with yo, y,, yn, ··· as a systemof
generators.PROOF: X can be obtained from the
wedge
of the twoloops generating
x
and y by adding
on asingle
2-cellaccording
to a map with thehomotopy
class
x- lyxy-2.
Hence the statement about x iswell-known,
see e.g.Spa-
nier
([3],
Theorem10,
p.147).
x is an element of infinite order in x. Let F be the infinite
cyclic
sub-group of 03C0
generated by x
and let C be thesubgroup
of 03C0generated by
the elements y0, y1, ···, yn, - - -.
Clearly
F n C containsonly
the neu-tral element. One proves
easily
that any word in x can beexpressed
inthe form
xmw,
where m is aninteger
and w is a word in C. Hence 7r isgenerated by
F and C.Altogether
03C0 is therefore a semidirectproduct
ofthe
subgroups
F and C.Using
the relationx-1yx
=y2
it is easyto prove that y2n+1
=yn, and
hence that yn =
(yn+k)2k’ Having
this information one verifies that C is abelian.Since non-trivial powers of x never commute
with y
=yo, it
is nowclear that
Cy(03C0)
= C.Clearly
C isinfinitely generated,
since one can neverproduce
the ele-ment yn+ 1 out of a
system
of éléments in Cinvolving only
yo , ’ ’ ’, yn . This provesProposition
2.PROPOSITION 3: X is an
Eilenberg-MacLane
spaceof
type(03C0, 1).
PROOF: Let X’ denote the
polyhedron
obtained from the space inFigure
1by identifying
this timeonly
the tworight-hand
circles. Con-sider then the double infinite
telescope
X" obtainedby glueing together copies
of X’ in bothends, always
such that the left-hand circle in any copy of X’ is identified with thepair
of identifiedright-hand
circles in theneigh- bouring
copy of X’ to the left in thetelescope.
Pushing
onestage
from the left to theright
in thetelescope
defines inthe obvious way an action of the
integers
Z on X". Thequotient
space for this action canclearly
be identified with X.Furthermore,
thequotient
36
map X" ~ X is a
covering projection.
To prove that X is anEilenberg-
MacLane space
of type (03C0, 1),
it suffices therefore to prove that03C0n(X") =
0 for
all n ~
2.For that purpose, let
f :
Sn ~ X" be anarbitrary
based map of then-sphere
Snfor n ~
2 into X".By
acompactness argument
theimage
ofthis map lies in a finite
stage
of thetelescope.
Therefore it is easy to see that we canhomotope f into
a map g : Sn ~X",
which maps one hemi-sphere
of ,Sn into along string,
and maps the otherhemisphere
of S" intoa circle
separating
twocopies
of X’sufficiently
far to theright
in thetelescope. Furthermore,
we can arrangethat g
maps the common boun-dary
of the twohemispheres
into asingle point.
It is then clearthat g,
and hence
also f,
ishomotopic
to the constant map. Thereforenn(X")
= 0for
all n ~ 2,
and asalready
remarked thiscompletes
theproof
of Pro-position
3.We are now
ready
for thePROOF OF THE THEoREm:
By Propositions
2 and3,
X is anaspherical
space with
finitely generated
fundamental group.Let 1;, : S1 ~
X be a based maprepresenting
thehomotopy
class[fy] =
YEn= 03C01(X,*). By Proposition 1, 03C01(XS1, fy)=Cy(03C0),
which is in-finitely generated by Proposition
2. Hence thecomponent
ofXS1,
whichcontains the
map fy,
hasinfinitely generated
fundamental group. Therefore X satisfies therequirements
in the theorem.REFERENCES
[1] H. FEDERER: A study of function spaces by spectral sequences. Trans. Amer.
Math. Soc. 82 (1956) 340-361.
[2] S. T. Hu: Homotopy Theory. Academic Press, 1959.
[3] E. H. SPANIER: Algebraic Topology. McGraw-Hill, 1966.
[4] R. THOM: L’homologie des espaces fonctionnels. Colloque de topologie algé- brique, Louvain 1956. Thone, Liège; Masson, Paris, 1957, pp. 29-39.
(Oblatum: 4-X-1972 & 1-VIII-1973) Matematisk Institut K0benhavns Universitet K0benhavn
Denmark